Jamie is walking along a number line that starts at point 0 and ends at point n. She can move either one step to the left or one step to the right of her current location , with the exception that she cannot move left from point 0 or right from point n. In other words, if Jamie is standing at point i,she can move to either i-1 or i+1 as long as her destination exists in the inclusive range [0,n]. She has a string ,s , of movement instruction consisting of the letters 1 and r , where 1 is an instruction to move one step left and r is an instruction to move one step right. Jamie followed the instructions in s one by one and in order .For Example if s=‘rrlr’,she performs the following sequence of moves :one step right ->one step right ->one step left -> one step right .Jamie wants to move from point x to point y following some subsequence of string s instruction and wonders how many distinct possible subsequence of string s will get her from point x to point y. recall that a subsequence of a string is obtained by deleting zero or more characters from string .

it has four parameters A String , s giving a sequence of eduction using the characters l( i.e. move left one unit ) and r (i.e. move right one unit) An integer n, denoting the length of the number line. An integer x, denoting jamie’s starting point on the number line An integer y , denoting Jamie’s enidng point on the number line. The function must return an integer denoting the total number of distinct subsequence of string s that will lead Jamie from point x to point y as this value cab be quite large .

Sample Input rrlrlr 6 1 2

out put =7

r
rrl
rlr
lrr
rrlrl
rlrlr
rrllr

Problem statement — Permutation is called stable if P[i]=i for every i

We are given Permutation we want to tell in how many moves we can make this permutation stable or print -1 if not possible to make permutation stable.

in one move we can do operation : — P[i] = P[P[i]]

Sample TestCase 
7
[1 3 2 5 6 7 4]
Output  - > 2

[1 3 2 5 6 7 4] -> [1 2 3 6 7 4 5] -> [1 2 3 4 5 6 7]

Sample TestCase 
7
[2 3 1 5 6 7 4]
Output  - > -1

Please help and thanks in advance.

Problem Link

In this problem,Nodes with duplicate values are not present.

class Solution {
    public boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if (root1 == root2)
            return true;
        if (root1 == null || root2 == null || root1.val != root2.val)
            return false;

        return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right) ||
                flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left));
    }
}

What would be the time complexity of this code if node with duplicate values were also present??

How Can I improve my code so that it also works with duplicate values?

Thanks in Advance:)

I am not able to point out my mistake in this code Can someone help me out to debug my code

Submission

I tried to solve this problem using centroid decomposition methed explained in this editorail :- editoial

Thanks in advance!.

Problem Link : https://www.interviewbit.com/problems/longest-common-subsequence/

Can anyone help me out.

why this code is giving time limit error(tle):

int fun(vector<vector<int> > &dp,int i,int j,string s1,string s2)
{
        if(i==s1.length()|| j==s2.length()) return 0;
        int &ans=dp[i][j];
        if(ans!=-1) return ans;
        ans=max(fun(dp,i,j+1,s1,s2),fun(dp,i+1,j,s1,s2));
        if(s1[i]==s2[j])
            ans=max(ans,1+fun(dp,i+1,j+1,s1,s2));
        return ans;
    }

 
int Solution::solve(string s1, string s2) {
    if(!s1.length() || !s2.length())    return 0;
    int n=s1.size(),m=s2.size();
    vector<vector<int> > dp(n,vector<int> (m,-1));
    return fun(dp,0,0,s1,s2);
}

and why this code get accepted:

×

int Solution::solve(string s1, string s2) {
    if(!s1.length() || !s2.length())    return 0;
    int n=s1.size(),m=s2.size(),i,j;
    vector<vector<int> > dp(n+1,vector<int> (m+1,0));
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
        {
            if(s1[i-1]==s2[j-1])
                dp[i][j]=1+dp[i-1][j-1];
            else
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
        }
    return dp[n][m];
}

both solutions have time complexity O(m*n) memoization gives tle and tabulation passes the constraints why?? Thanks in advance.